Question: Perform the row operation, $R_1+4R_3\rightarrow R_1$, on the following matrix. $\left[\begin{array} {ccc} -3 & -4 & 3 & 0 \\ 9 & 7 & 1 & 2 \\ 2 & 8 & 6 & 5 \end{array} \right] $
Solution: Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_1$ and $R_3$ are given below. $R_1=\left[\begin{array} {ccc} -3 & -4 & 3 & 0 \end{array} \right] ~~~~~ R_3=\left[\begin{array} {ccc} 2 & 8 & 6 & 5 \end{array} \right]$ We are asked to perform the row operation, $R_1+4R_3\rightarrow R_1$. Therefore, we must add $4R_3$ to $R_1$. $\begin{aligned}R_1+4R_3 &= \left[\begin{array} {ccc} -3 & -4 & 3 & 0 \end{array} \right] + 4\left[\begin{array} {ccc} 2 & 8 & 6 & 5 \end{array} \right] \\\\&=\left[\begin{array} {ccc} 5 & 28 & 27 & 20 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_1$ with $R_1+4R_3$. $\left[\begin{array} {ccc} {-3} & {-4} & {3} & {0} \\ 9 & 7 & 1 & 2 \\ 2 & 8 & 6 & 5 \end{array} \right] \xrightarrow{R_1+4R_3\rightarrow R_1} \left[\begin{array} {ccc} {5} & {28} & {27} & {20} \\ 9 & 7 & 1 & 2 \\ 2 & 8 & 6 & 5 \end{array} \right] $ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} 5 & 28 & 27 & 20 \\ 9 & 7 & 1 & 2 \\ 2 & 8 & 6 & 5 \end{array} \right]$